$f(x, y, z) = (x^4 - y^2, z^2 + 3x^4, 2y^2 + 2)$ $\text{curl}(f) = $ $\hat{\imath} + $ $\hat{\jmath} + $ $\hat{k}$
Answer: $f(x, y, z) = (f_0, f_1, f_2)$ The curl of $f$ : $\begin{aligned} \text{curl}(f) &= \det \begin{bmatrix} {\hat{\imath}} & \hat{\jmath} & \hat{k} \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ f_0 & f_1 & f_2 \end{bmatrix} \\ \\ &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \end{aligned}$ $\begin{aligned} f_0(x, y, z) &= x^4 - y^2 \\ \\ f_1(x, y, z) &= z^2 + 3x^4 \\ \\ f_2(x, y, z) &= 2y^2 + 2 \end{aligned}$ Let's calculate all the partial derivatives we'll need. $f_0$ $f_1$ $f_2$ $\dfrac{\partial}{\partial x}$ $12x^3$ $0$ $\dfrac{\partial}{\partial y}$ $-2y$ $4y$ $\dfrac{\partial}{\partial z}$ $0$ $2z$ Now we can put it all together. $\begin{aligned} \text{curl}(f) &= \left( \dfrac{\partial f_2}{\partial y} - \dfrac{\partial f_1}{\partial z} \right) \hat{\imath} \\ \\ &+ \left( \dfrac{\partial f_0}{\partial z} - \dfrac{\partial f_2}{\partial x} \right) \hat{\jmath} \\ \\ &+ \left( \dfrac{\partial f_1}{\partial x} - \dfrac{\partial f_0}{\partial y} \right) \hat{k} \\ \\ &= (4y - 2z) \hat{\imath} + (0 - 0) \hat{\jmath} + (12x^3 + 2y) \hat{k} \\ \\ &= (4y - 2z) \hat{\imath} + 0 \hat{\jmath} + (12x^3 + 2y) \hat{k} \end{aligned}$ In conclusion: $\text{curl}(f) = (4y - 2z) \hat{\imath} + 0 \hat{\jmath} + (12x^3 + 2y) \hat{k}$